This reaction involves the hydration of an alkyne (H₃C–CH₂–C≡C–OEt) using Hg²⁺/H⁺/H₂O. This is Kucherov's reaction, which adds water across the triple bond following Markovnikov's rule. The terminal alkyne forms an enol that tautomerizes to a ketone. The –OEt (ethoxy) group is an electron-withdrawing substituent but does not participate in the hydration; the reaction occurs at the triple bond. The product Q is a ketone where the carbonyl group is on the carbon that was the terminal carbon of the alkyne.

The final structure of Q is: $H_{3}C-{\mathrm{CH}}_2-C(=O)-C{H}_2-O\mathrm{Et}$

Final Answer: The third option.